For positive $a$, $b$, $c$ with $abc=1$, show that $\sum_{cyc}\sqrt{a^2-a+1}\geq a+b+c$ 
Let $a,b,c$ are positive number such that $abc=1$. Prove that:
  $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\;\geq\; a+b+c$$

This problem froms my Math teacher. I have attempted to let $$(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$. The inequality is equivalent to: $\frac{\sqrt{x^2-xy+y^2}}{y}+\frac{\sqrt{y^2-yz+z^2}}{z}+\frac{\sqrt{x^2-xz+z^2}}{x}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$
Then, I tried to use AM-GM but I stucked on it.
 A: The hint:
Use the Mixing Variables method.
Indeed, we can use the beautiful Can's idea.
Since $$\prod\limits_{cyc}(a-1)^2=\prod_{cyc}((a-1)(b-1))\geq0,$$ we can assume that $$(a-1)(b-1)\geq0$$ or
$$a+b\leq1+ab=1+\frac{1}{c}.$$
Thus, by C-S:
\begin{align}
\sqrt{a^2-a+1}&+\sqrt{b^2-b+1}\\
&=\sqrt{a^2+b^2-a-b+2+2\sqrt{(a^2-a+1)(b^2-b+1)}}\\
&\geq\sqrt{a^2+b^2-a-b+2+2\sqrt{\left(\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right)\left(\left(b-\frac{1}{2}\right)^2+\frac{3}{4}\right)}}\\
&\geq\sqrt{a^2+b^2-a-b+2+2\left(\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)+\frac{3}{4}\right)}\\
&=\sqrt{a^2+b^2-a-b+2+2ab-a-b+2}\\
&=\sqrt{(a+b)^2-2(a+b)+4}.
\end{align}
But $f(x)=\sqrt{x^2-2x+4}-x$ decreases, which says
$$\sum_{cyc}(\sqrt{a^2-a+1}-a)\geq f(a+b)+\sqrt{c^2-c+1}-c\geq$$
$$\geq f\left(1+\frac{1}{c}\right)+\sqrt{c^2-c+1}-c=\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c.$$
Id est, it's enough to prove that:
$$\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c\geq0$$ and the rest is smooth.
Can you end it now?
